Problem: What is the slope of the line tangent to $f(x) = -2x^{2}+4x+5$ at $x = -1$ ?
Solution: The slope of the tangent line is $ \lim_{\Delta x \to 0} \frac{f(x + \Delta x) - f(x)}{\Delta x}$ $ = \lim_{\Delta x \to 0} \frac{(-2(x+\Delta x)^{2}+4(x+\Delta x)+5) - (-2x^{2}+4x+5)}{\Delta x}$ $ = \lim_{\Delta x \to 0} \frac{(-2(x^{2}+2x \Delta x+\Delta x^{2})+4(x+\Delta x)+5) - (-2x^{2}+4x+5)}{\Delta x}$ $ = \lim_{\Delta x \to 0} \frac{-2x^{2}-4(x \Delta x)-2\Delta x^{2}+4x+4(\Delta x)+5+2x^{2}-4x-5}{\Delta x}$ $ = \lim_{\Delta x \to 0} \frac{-4(x \Delta x)-2\Delta x^{2}+4(\Delta x)}{\Delta x}$ $ = \lim_{\Delta x \to 0} -4x-2(\Delta x)+4$ $ = -4x+4$ $ = (-4)(-1)+4$ $ = 8$